Question

Помогите прошу вас сроооочно

Answer 1

Ответ:

$1)\ \ a_{n}=n^2-12n+17\\\\n=-\dfrac{-12}{2}=6\\\\...\ ,\ a_4=-15\ ,\ \ a_5=-18\ \ ,\ \ a_6=-19\ \ ,\ \ \ a_7=-18\ ,\ \ a_8=-15\ ,...\\\\a_{naimenshee}=a_6=-19$

$2)\ \ y_4-y_2=-24\ \ ,\ \ y_3+y_2=6\ \ ,\ \ \ S_{n}=-182\\\\y_4-y_2=y_1q^3-y_1q=y_1q\, (q^2-1)\ \ ,\ \ \ y_1q\, (q^2-1)=-24\\\\y_3+y_2=y_1q^2+y_1q=y_1q\, (q+1)\ \ ,\ \ \ \ y_1q\, (q+1)=6\\\\\\\dfrac{y_4-y_2}{y_3+y_2}=\dfrac{y_1q\, (q^2-1)}{y_1q\, (q+1)}=\dfrac{q^2-1}{q+1}=\dfrac{(q-1)(q+1)}{q+1}=q-1\\\\\\q-1=\dfrac{-24}{6}\ \ \ ,\ \ \ q-1=-4\ \ \ \to \ \ \ \boxed{q=-3}\\\\\\6=y_1q\, (q+1)\ \ ,\ \ \ 6=y_1\cdot (-3)\cdot (\underbrace{-3+1}_{-2})\ \ ,\ \ \ 6=6y_1\ \ ,\ \ \ \boxed{y_1=1}$

$S_{n}=-182\ \ ,\ \ S_{n}=\dfrac{y_1\, (q^{n}-1)}{q-1}\ \ ,\ \ \ \ -182=\dfrac{1\cdot ((-3)^{n}-1)}{-3-1}\ \ ,\\\\\\\dfrac{(-3)^{n}-1}{-4}=-182\ \ ,\ \ \ (-3)^{n}-1=728\ \ ,\ \ \ (-3)^{n}=729\ \ ,\ \ (-3)^{n}=(-3)^6\ ,\\\\\\n=6\ \ \ \to \ \ \ \ S_6=-182\\\\\boxed{\ \{y_{n}\}:\ \ \ 1\ ;\ -3\ ;\ 9\ ;\ -27\ ;\ 81\ ;\ -243\ ;\ ...}$

$3)\ \ \ \{x_{n}\}:\ \ ...\ ;\ a\ ;\ b\ ;\ c\ ;\ ...\ -\ arifm.\ progressiya\ \ ,\ \ (a+2b)^2=8ab+c^2\\\\x_{n}=a\ \ ,\ \ x_{n+1}=b=a+d\ \ ,\ \ \ x_{n+2}=c=b+d=a+2d\\\\\\\underline {(a+2b)^2}=(a+2a+2d)^2=(3a+2d)^2=\underline {9a^2+12ad+4d^2}\ ;\\\\\underline{8ab+c^2}=8a\cdot (a+d)+(a+2d)^2=8a^2+8ad+a^2+4ad+4d^2=\\\\=\underline{9a^2+12ad+4d^2}\ ;\\\\\\\boxed{\ (a+2d)^2=8ab+c^2\ }$